A particle of mass 0.5kg in a straight line with velocity v=3x^3/2 where a=5m^-1/2 s^-1/2. What is the work done by the net force during the displacement from x=0 from x=2 m.
Answers
Answer:
Velocity of body is given as v=ax
3/2
where a=5
Initial velocity of body v
i
∣
∣
∣
∣
∣
x=0
=5×0=0 m/s
Final velocity of body v
f
∣
∣
∣
∣
∣
x=2
=5×2
3/2
=10
2
m/s
From work-energy theorem, work done by the force is equal to change in kinetic energy of the body.
∴ W=
2
1
mv
f
2
−
2
1
mv
i
2
Or W=
2
1
(0.5)(10
2
)
2
−
2
1
(0.5)(0)
Or W=
2
1
(0.5)(200)−0
⟹ W=50 J
Step-by-step explanation:
Answer:
50J is answer mate please mark me as brainliest
Step-by-step explanation:
Given, v=ax3/2
We know that
Acceleration
a0=dv/dt=v dv dx=ax3/2ddx(ax3/2)
=ax3/2×a×32×x1/2=32a2x2
Now, Force =ma0=m32a2x2
Work done =∫x=2x=0Fdx=∫2032ma2x2dx
=32ma2×(x3/3)20
=12ma2×8=12×(0.5)×(25)×8=50J